If the 8th term of an A.P. is 31 and the 15th term is 16 more than its 11th term, then find the A.P.

Given, a₈ = 31 and a₁₅ - a₁₁ = 16.

We know that

⇒ a_n = a + (n - 1)d
∴ a₈ = a + 7d      (Eq 1)
   a₁₅ = a + 14d and
   a₁₁ = a + 10d.

Given, a₁₅ - a₁₁ = 16.
∴ a + 14d - (a + 10d) = 16
⇒ a - a + 14d - 10d = 16
⇒ 4d = 16
⇒ d = 4.

Putting value of d in Eq 1
⇒ a₈ = a + 7d
⇒ a + 7 × 4 = 31
⇒ a + 28 = 31
⇒ a = 31 - 28
⇒ a = 3.

Hence, the terms of A.P. are

a₂ = a₁ + d = 3 + 4 = 7,
a₃ = a₂ + d = 7 + 4 = 11,
a₄ = a₃ + d = 11 + 4 = 15.

Hence, the terms of the A.P. are 3, 7, 11, 15, ….